JEE (Main) 2015 Question Paper with Solution Chemistry · Which of the following compounds will...

1JEE (Main) 2015 Question Paper with Solution

JEE (Main) 2015 Question Paper with Solution

Chemistry

* Questions from XIth Syllabus

1.* Which of the following is the energy of a possible excited state of hydrogen ?(1) –6.8 eV (2) –3.4 eV (3) +6.8 eV (4) +13.6 eV .

Sol.[2] n 2

13.6E

n

eV

Excited state when n = 2, 3, 4-------n = 2, E

2 = –3.4 ev.

2. In the following sequence of reactions :

, the product C is :

(1) C6H

5CH

3 (2) C

6H

5CH

2OH (3) C

6H

5CHO (4) C

6H

5COOH

Sol.[3]

3.* Which compound would give 5 – keto – 2 – methyl hexanal upon ozonolysis ?

(1) (2) (3) (4)

Sol.[1]

5- Keto-2-methyl hexanal

4.* The ionic radii (in Å) of N3–, O2 – and F– are respectively :(1) 1.36, 1.71 and 1.40 (2) 1.71, 1.40 and 1.36 (3) 1.71, 1.36 and 1.40 (4) 1.36, 1.40 and 1.71

Sol.[2] For isoelectronic species, Size of ion increases with increase of –ve charge

3 2N O F order of size

2 JEE (Main) 2015 Question Paper with Solution

5. The color of KMnO4 is due to :

(1) d – d transition (2) L M charge transfer transition(3) – * transition (4) M L charge transfer transition

Sol.[2] Color of KMnO4 is due to charge transfer from O2– (Ligand) to Mn (VII) (Central Metal ion ).

6. Assertion : Nitrogen and Oxygen are the main components in the atmosphere but these do not react to formoxides of nitrogen.Reason : The reaction between nitrogen and oxygen requires high temperature.(1) Both assertion and reason are correct, but the reason is not the correct explanation for the assertion(2) The assertion is incorrect, but the reason is correct(3) Both the assertion and reason are incorrect(4) Both assertion and reason are correct, and the reason is the correct explanation for the assertion

Sol.[4] Oxides of nitrogen is formed at 20000C such as during lightening.

7. Which of the following compounds is not an antacid ?(1) Cimetidine (2) Phenelzine (3) Ranitidine (4) Aluminium hydroxide

Sol.[2] Phenelzine is anti depressant

8. In the context of the Hall – Heroult process for the extraction of Al, which of the following statements is false ?(1) Al

2O

3 is mixed with CaF

2 which lowers the melting point of the mixture and brings conductivity

(2) Al3+ is reduced at the cathode to form Al(3) Na

3AlF

6 serves as the electrolyte

(4) CO and CO2 are produced in this process

Sol.[3] Read NCERT chapter 6 Part (XII) at page no. 158.(1) In this process, carbon anode is oxidised to CO and CO

2.

(2) it is fact(3) At cathode, Al3+ from Al

2O

3 is reduced to Al.

(4) Al2O

3 serves as electrolyte and Na

3AlF

6 serves as solvent.

9. Match the catalysts to the correct processes :Catalyst Process(A) TiCl

3(i) Wacker process

(B) PdCl2

(ii) Ziegler – Natta polymerization(C) CuCl

2(iii) Contact process

(D) V2O

5(iv) Deacon's process

(1) (A) – (ii), (B) – (i), (C) – (iv), (D) – (iii) (2) (A) – (ii), (B) – (iii), (C) – (iv), (D) – (i)(5) (A) – (iii), (B) – (i), (C) – (ii), (D) – (iv) (4) (A) – (iii), (B) – (ii), (C) – (iv), (D) – (i)

Sol.[1] TiCl3 is Zieglar Natta catalyst & V

2 O

5 is used in contact process.

PdCl2 is catalyst of Wackur process.

CuCl2 is catalyst of Deacon’s process.

10. In the reaction

the product E is :

(1) (2) (3) (4)


Sol.[2]

11. Which polymer is used in the manufacture of paints and lacquers ?(1) Glyptal (2) Polypropene (3) Poly vinyl chloride (4) Bakelite

Sol.[1] Glyptal is used in paints & lacquer.

12. The number of geometric isomers that can exist for square planar [Pt (Cl) (py) (NH3) (NH

2OH)]+ is (py = pyridine):

(1) 3 (2) 4 (3) 6 (4) 2

Sol.[1] 3 geometrical isomers.

13. Higher order (>3) reactions are rare due to :(1) increase in entropy and activation energy as more molecules are involved(2) shifting of equilibrium towards reactants due to elastic collisions(3) loss of active species on collision(4) low probability of simultaneous collision of all the reacting species

Sol.[4] Low probability of simultaneous collision of all the reacting species.

14. Which among the following is the most reactive ?(1) Br

2(2)

2(3) Cl (4) Cl

2

Sol.[3] I+– Cl– is a polar molecule. So, interhalogen compounds are more reactive.

15. Two Faraday of electricity is passed through a solution of CuSO4 . The mass of copper deposited at the cathode

is : (at. mass of Cu =63.5 amu)(1) 63.5 g (2) 2 g (2) 127 g (4) 0 g

Sol.[1] Cu2+ +2e– Cu2 Faraday of electricity will deposit 1 mole of Cu= 63.5 g

16. 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filteredand the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal)is :(1) 36 mg (2) 42 mg (3) 54 mg (4) 18 mg

Sol.[4] Mass of acetic acid adsorbed= 50×(0.06– 0.042)×62 mgAmount of acetic acid adsorbed(Per gram of Charcoal)

= 50 0.06 0.042 62

3

= 18.6 mg

17. The synthesis of alkyl fluorides is best accomplished by :

(1) Sandmeyer's reaction (2) Finkelstein reaction (3) Swarts reaction (4) Free radical fluorination


Sol.[3] Read NCERT (XII) Page No. 289

R–X+AgFR–F+AgX (Swarts reaction)

18. The molecular formula of a commercial resin used for exchanging ions in water softening is C8H

7SO

3Na (Mol. wt.

206). What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin?

(1) 1

206(2)

2

309(3)

1

412(4)

1

103

Sol.[3] 2C8H

7SO

3Na+Ca2+ C

8H

7SO

3

2Ca + 2Na+

2 mole is reacting with 1 mole of Ca2+ ions. Maximum uptake of Ca2+ ions by the resin when expressed in

moles per gram of resin = 1 1

2 206 412

.

19. Which of the vitamins given below is water soluble ?

(1) Vitamin D (2) Vitamin E (3) Vitamin K (4) Vitamin C

Sol.[4] Vitamin C & Vitamin B are water soluble.

20.* The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is :

(1) ion – dipole interaction (2) London force (3) hydrogen bond (4) ion – ion interaction

Sol.[3] Read NCERT Part I (XI) Chapter -5 page No. 133

21.* The following reaction is performed at 298 K. 2NO(g)+ O2(g) 2NO

2(g). The standard free energy of formation of

NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2(g) at 298 K? (K

p=1.6×1012)

(1) 86600 + R(298) In(1.6×1012) (2) 86600 –

12In 1.6 10

R 298

(3) 0.5[2×86,600 – R(298) In(1.6×1012) (4) R(298) In(1.6×1012) – 86600

Sol.[3] 2 22NO g O g 2NO g

0 0F,NO F,No2

0reaction 2G 2G 1G

Also, p

0reaction 2.303RT logK 2G

From (1) & (2)

–2.303 RT log Kp = 2G0

F NO

2– 2G0

p, NO

0 0F F p

1G , = 2G , NO 2.303 R.T. log K

2

= 1212 86600 2.303R 298 log1.6 10

2

22.* Which of the following compounds is not colored yellow ?

(1) K3[Co(NO

2)6] (2) (NH

4)

3 [As (Mo

3 O

10)

4] (3) BaCrO

4(4) Zn

2[Fe(CN)

6]

Sol.[4] 2 6Zn Fe CN

is not yellow color instead white color..

23.* In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentageof bromine in the compound.is :


(at. mass Ag =108; Br =80)(1) 36 (2) 48 (3) 60 (4) 24

Sol.[4] All Br from organic compound is coming to AgBr.188 g of AgBr contains 80g of Br

So, 141 mg of AgBr contains = 80

141mg188

% of Br = 80 141

100 24%188 250

24. Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29Å. The radius of sodium atomis approximately : (1) 3.22Å (2) 5.72Å (3) 0.93Å (4) 1.86Å

Sol.[4] In BCC, a 3 4r

r = 04.29 31.86A

4

25.* Which of the following compounds will exhibit geometrical isomerism ?(1) 3 – Phenyl – 1 – butene (2) 2 – Phenyl – 1 – butene(3) 1, 1 – Diphenyl – 1 – propane (4) 1 – Phenyl – 2 – butene

Sol.[4] 2 3PhCH CH CH CH will show geometrical isomerism.

cis trans

26. The vapour pressure of acetone at 20°C is 185 torr. When 1.2 g of a non–volatile substance was dissolved in 100g of acetone at 20°C, its vapour pressure was 183 torr. The molar mass (g mol–1) of the substance is:(1) 64 (2) 128 (3) 488 (4) 32

Sol.[1] 0P Ps n 185 183 1.2MPs N 183

100

62

wM 64

27.* From the following statements regarding H2O

2, choose the incorrect statement :

(1) It decomposes on exposure to light(2) It has to be stored in plastic or wax lined glass bottles in dark(3) It has to be kept away from dust(4) It can act only as an oxidizing agent

Sol.[4] 2 2H O acts both reducing and oxidising agent because O is in –1 oxidation state which is its intermediate

state.

28.* Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its latticeenthalpy ?(1) BeSO

4(2) BaSO

4(3) SrSO

4(4) CaSO

4

Sol.[1] Solubility order

4 4 4 4BeSO CaSO SrSO BaSO


29.* The standard Gibbs energy change at 300 K for the reaction 2A B + C is 2494.2 J. At a given time, the

composition of the reaction mixture is [A] = 1

2, [B]=2 and [C]=

1

2. The reaction proceeds in the : [R =8.314 J/K/

mol, e=2.718]

(1) reverse direction because Q > Kc

(2) forward direction because Q < Kc

(3) reverse direction because Q < Kc

(4) forward direction because Q > Kc

Sol.[1] 2A B C

2 2

12B C 2Q 41A

2

0eqG 2.303RT log K

2494.2 = –2.303×8.314×300 log eqK

eq2494.2

logK2.303 8.314 300

eqlogK 0.434 1 0.566

1eqK antilog 0.566 10

= 0.368

Q > eqK

So, reaction will go in backward direction

30.* Which one has the highest boiling point ?(1) Ne (2) Kr (3) Xe (4) He

Sol.[3] Xe Kr Ne He Boiling point

In this case, Boiling point order depends on London forces of attraction which depends on the molecular massof the compounds.


Mathematics

31.* The sum of coefficients of integral powers of x in the binomial expansion of is

(1) (2) (3) (4)

Sol.[4] 1 2 x 50 1 2 x 50

50 50 2 50 2 50 50 252 2 4 502 C C 2 x C 4x C 2 x

Put x = 1

5050 50 2 50 4 50 50

0 2 4 501 3

C C 2 C 2 C 22

32. Let f(x) be a polynomial of degree four having extreme values at x=1 and x=2.

2x 0

f xlim 1

x

=3, then f (2) is equal

to:(1) – 4 (2) 0 (3) 4 (4) – 8

Sol.[2] 2f x ax x 1 x 2 as 2x 0

f xlim

x exists and is finite.

2x 0

f xlim

x = 2a = 2 a= 1

2f x x x 1 x 2 f 2 0

33.* The mean of the data set comprising of 16 observations is16. If one of the observation valued 16 is deleted andthree new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is :(1) 16.0 (2) 15.8 (3) 14.0 (4) 16.8

Sol.[3]i

ix

x 16 x 16 1616

New mean = 1x 16 3 4 5 16 16 4

1418 18

34.* The sum of first 9 terms of the series is

(1) 96 (2) 142 (3) 192 (4) 71

Sol.[1]

3

nn

T1 3 5 2n 1

2

2

2

n n 1

2 n 1

4n

2n

1S n 1 2n

4

n

n n 1 2n 1 n n 11S b 2

4 6 2

99 10.19

s 1 10 964 6


2=8y. If the point P divides the line segment OQ internallyin the ratio 1 : 3, then the locus of P is :(1) y2=x (2) y2=2x (3) x2=2y (4) x2=y

Sol.[3]

1 4t 3 0h t

4

21 2t 3 0k

4

=

22 2t

2k h x 2y2

36.* Let and be the roots of equation x2–6x–2=0. If an=n–n, for n > 1, then the value of

10 8

9

a 2a

2a

is equal to :

(1) –6 (2) 3 (3) –3 (4) 6

Sol.[2]

10 10 8 8

10 89 9

9

2a 2a

2a 2

8 2 8 2

9 9

2 2

2

8 8

9 9

6 6

2

6

32

(using 2 2 6 )

37. If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes containsexactly 3 balls is:

(1)

102

553

(2)

121

2203

(3)

111

223

(4)

1155 2

3 3

Sol.[*] When identical balls are to be distributed into identical boxes, different cases need to be taken e.g. for 3 ballsinto 3 boxes ; (3, 0, 0), (2, 1, 0), (1, 1, 1). Here, the number of balls being 12, points to huge number of cases,hence the question involves theoretical error.

38. A complex number z is said to be unimodular if |z|=1. Suppose z1 and z

2 are complex numbers such that

1 2

1 2

z 2z

2 z z

is unimodular and z2 is not unimodular. Then the point z

1 lies on a :

(1) straight line parallel to y–axis. (2) circle of radius 2.

(3) circle of radius 2 . (4) straight line parallel to x–axis,

Sol.[2]1 2

21 2

z 2z1 & z 1

2 z z

1 2 1 2z 2z 2 z z

Let 2 2 21 1 1 2 2 2 2 2 2z x iy , z x iy , z x y 1


21 2 1 2x 2x i y 2y 2

1 1 2 22 x iy x iy

2 2 2 2 2 21 1 2 2 2 2x +y 1 x y 4 1 x y

2 21 1 1 1x y 4 x , y lies on circle of radius 2.

39. The integral equals : 3

42 4

dx

x x 1 equals

(1) 1

4 4x 1 c (2) 1

4 4x 1 c (3)

14 4

4

x 1c

x

(4)

14 4

4

x 1c

x

Sol.[3] 3

2 3 4 4

dx

x . x 1 x

=

5

34 4

x dx

1 x

Put 1+x–4 = t4 –4x–5 dx = 4t3 dt

= 13

4 43

t dtdt t c 1 x c

t

40.* The number of points, having both co–ordinates as integers, that lie in the interior of the triangle with vertices(0, 0), (0, 41) and (41, 0), is :

(1) 861 (2) 820 (3) 780 (4) 901

Sol.[3]

Points inside the triangle are

(1, 1), (1, 2)------ (1, 39) = 39

(2, 1), (2, 2)------ (2, 38) = 38

.

.

(39, 1) = 1

Hence number of points are 39+38+37.......+1

=39.40

2 = 780

Alt :

x+y < 41 x+y+z= 41, where x, y, z>1

No. of points = No. of ways 41 coins can be distributed among 3 beggars each getting atleast

one coin = 38 22C


41. The distance of the point (1, 0, 2) from the point of intersection of the line and the plane x–

y+z=16, is :

(1) 8 (2) 3 21 (3) 13 (4) 2 14

Sol.[3] x=3+2, y= 4–1, z= 12+2For point of intersection with the plane3+2– (4–1)+ (12+2) = 16

=1 ( 5, 3, 14) 2 2 2d 4 3 12 13

42. The equation of the plane containing the line 2x–5y+z=3; x+y+4z=5, and parallel to the plane, x+3y+6z=1, is :(1) x+3y+6z= –7 (2) x+3y+6z=7 (3) 2x+6y+12z=–13 (4) 2x+6y+12z=13

Sol.[2] Put z= 0 : 2x–5y = 3 x+y = 57y = 7 y=1 x=4 The plane passes through (4, 1, 0)x+3y+6z=4+3(1)+6 (0) = 7

43. The area (in sq. units) of the region described by {(x, y): y2 < 2x and y > 4x – 1} is:

(1) 5

64(2)

15

64(3)

9

32(4)

7

32

Sol.[3] Using Horizontal strip,

1 2

1

2

y 1 y 9A dy

4 2 32

44.* If m is the A.M. of two distinct real numbers l and n (l, n > 1) and G1, G

2 and G

3 are three geometric means between

l and n, then G 41 + 2G 4

2 + G 43 equals.

(1) 4 lm2n (2) 4 lmn2 (3) 4 l2m2n2 (4) 4 l2mn

Sol.[1] 2m n

1 2 3, G ,G ,G ,n are in G.P..

Common ratio

14n

r

2 34 4 4 41 2 3

n n nG 2G G 2

= 2 2n n 4 m n

45.* Locus of the image of the point (2, 3) in the line (2x –3y+4)+k (x–2y+3)=0, k R, is a :

(1) straight line parallel to y–axis. (2) circle of radius 2

(3) circle of radius 3 (4) straight line parallel to x–axis.

Sol.[2]


The given equation represents family of lines passing through (1, 2). The locus of R is a circle with (1, 2) &(2, 3) as end- points of diameter.

h 2 h 2 k 3 k 31 2 2 3 0

2 2 2 2

which gives a circle of radius 2

46.* The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the

ellipse2 2x y

19 5

, is:

(1) 18 (2) 27

2(3) 27 (4)

27

4

Sol.[3] Area of the rhombus = 22a 9

2 27e 2 / 3

47.* The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition,is :(1) 192 (2) 120 (3) 72 (4) 216

Sol.[1] No. of 5 digit number = 5! = 120No. of 4 digit numbers = 3.4.3.2 = 72Hence, total numbers = 120+72=192

48.* Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A×B,each having at least three elements is :(1) 256 (2) 275 (3) 510 (4) 219

Sol.[4] n A B 8

No. of subsets having atmost two elements = 8 8 80 1 2C C C 37

No. of subsets having atleast three elements = 82 32 219

49. Let tan–1 y = tan–1x + tan–1 2

2x,

1 x

where |x| <

1

3. Then a value of y is :

(1) 3

2

3x + x

1 3x(2)

3

2

3x x

1+3x

(3)

3

2

3x +x

1+3x(4)

3

2

3x x

1 3x

Sol.[4] 1 1 1tan y tan x 2tan x

3

12

3x xy tan 3 tan x

1 3x


50. The integral 4 2

2 22

logxdx

logx log 36 12x x is equal to :

(1) 4 (2) 1 (3) 6 (4) 2

Sol.[2] I =

24

22 2

logxdx

logx log 6 x

24

22 2

logxdx

logx log 6 x

I

24

22 2

log 6 xdx

log 6 x logx

2 I4

2dx 2 I 1

51.* The negation of ~ s v (~ r s ) is equivalent to :(1) s (r ~ s) (2) s (r ~ s) (3) s r (4) s ~ r

Sol.[3] ~(~ s (~ r s))= s (r ~s)=(s r) (s ~s)=s r

52. If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to the foot ofthe tower, are 30°, 45° and 60° respectively, then the ratio, AB : BC, is :

(1) 3 : 2 (2) 1 : 3 (3) 2 : 3 (4) 3 : 1

Sol.[4]0 0h h

tan60 , tan 45x BC x

h 3x , BC x 3x BC 3 1 X

0 htan30

AB BC x

AB 3 1 x x 3x

AB = 3 3 x 3 3 1 x

AB 3

BC 1

53. lim (1 – cos 2x)(3 + cos x ) is equal to :x0 x tan 4x

(1) 3 (2) 2 (3) 1

2(4) 4

Sol.[2]

x 0

1 cos2x 3 cos xlim

x tan4x

2

x 0 x 0

2sin xlim lim 3 cos x

tan4xx .4x

4x

=1

2

2

x 0

x 0

sinxlim

x

tan4xlim

4x

4


a , b and c

be three non–zero vectors such that no two of them are collinear and 1

(a x b)×c= b c a3

. If is the

angle between vectors b and c

, then a value of sin is :

(1) 2

3

(2)

2

3(3)

2 3

3

(4)

2 2

3

Sol.[4] 1a.c b b.c a b c a

3

1b.c b c

3 a & b are non-collinear

1 2 2cos sin 0,

3 3

55. If A =

1 2 2

2 1 2

a 2 b

is a matrix satisfying the equation AAT=9I, where is 3×3 identity matrix, then the ordered pair

(a, b) is equal to :

(1) (– 2, 1) (2) (2, 1) (3) (–2, –1) (4) (2, –1)

Sol.[3]T

T

1 2 2

3 3 3A A A 2 1 2

AA 9I I3 3 3 3 3 3

a 2 b

3 3 3

is orthogonal matrix.

2a 2 2b0

9 9 9 &

a 4 2b0

9 9 9

a 2 & b 1

56. If the function g(x ) = k x 1 , 0 x 3

mx 2 , 3 x 5

is differentiable, then the value of k+m is :

(1) 16

5(2)

10

3(3) 4 (4) 2

Sol.[4] k x 1 , 0 x 3g x

mx 2 ,3 x 5

for continuity k 3 1 3m 2 2k 3m 2 ------(1)

k

, 0 x<3g x 2 x 1

m , 3 x 5

k

m4 ------(2)

(1) & (2) gives m k 2


57. The set of all values of for which the system of linear equations :2x

1–2x

2+x

3 x

1

2x1 –3x

2+2x

3= x

2

–x1+2x

2 = x

3

has a non–trivial solution,(1) is a singleton (2) contains two elements(3) contains more than two elements (4) is an empty set.

Sol.[2] 1 2 3 12x 2x x x

1 2 3 22x 3x 2x x

1 2 3x 2x x

For non-trivial solution

2 2 1

2 3 2 0

1 2

1, 3

58. The normal to the curve, x2+2xy –3y2=0, at (1, 1) :(1) meets the curve again in the second quadrant. (2) meets the curve again in the third quadrant.(3) meets the curve again in the fourth quadrant. (4) does not meet the curve again.

Sol.[3] 2 2x 2xy 3y 0

x y x 3y 0 x y 0 or x+3y=0

Normal at (1, 1) to the straight line y = x is : y–1 = –1 (x–1)x+y=2 which cuts the straight line x+3=0 at (3, –1).

59.* The number of common tangents to the circles x2+y2– 4x–6y–12=0 and x2+y2+6x+18y+26=0, is :(1) 2 (2) 3 (3) 4 (4) 1

Sol.[2] 2 21S : x y 4x 6y 12 0 ,

2 2x 2 y 3 25

2 22S : x y 6x 18y 26 0

2 2x 3 y 9 64

1 2r 5, r 8

distance between centres 1 2C C 13

C1C

2 = r

1 + r

2 S

1 & S

2 touch each other.

Hence there are three common tangents.

60. Let y(x) be the solution of the differential equation (x logx) dy

dx + y =2x logx, (x > 1). Then y(e) is equal to :

(1) 0 (2) 2 (3) 2e (4) e

Sol.[2] Dividing by x, dy 1nx .y 2 nx

dx x (at x = 1, y=0)

n x y 2 n x dx=2 x n x-x c

Using y 1 0, c=2

y e 2


Physics

61. As an electron makes a transition from an excited state to the ground state of a hydrogen – like atom/ion :

(1) kinetic energy, potential energy and total energy decrease(2) kinetic energy decreases, potential energy increases but total energy remains same(3) kinetic energy and total energy decrease but potential energy increases(4) its kinetic energy increases but potential energy and total energy decrease

Sol.[4] When e– jump from higher to lower state P.E. and T.E. K.E.

62.* The period of oscillation of a simple pendulum is T = 2 L

g . Measured value of L is 20.0 cm known to 1 mm

accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resolution. Theaccuracy in the determination of g is :(1) 3% (2) 1% (3) 5 % (4) 2%

Sol.[1] T 2ng

T 1 1 g100 100

T 2 2 g

g100 3%

g

63. A long cylindrical shell carries positive surface charge in the upper half and negative surface charge – in thelower half. The electric field lines around the cylinder will look like figure given in : (figures are schematic and notdrawn to scale)

(1) (2)

(3) (4)

Sol.[4]

64. A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of theresultant signal is/are :(1) 2005 kHz, and 1995 kHz (2) 2005 kHz, 2000 kHz and 1995 kHz(3) 2000 kHz and 1995 kHz (4) 2 MHz only

Sol.[2] In frequency modulation the resultant eqnation of =Vc cos

ct+

0mV

2 mcos t

+0mV

2 mcos t

So possible frequency is 2005 KHz., 2000 KHz and 1995 KHz

65.* Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an

ideal gas of photons with internal energy per unit volume u= 4U

TV

and pressure p =

1 U

3 V

. If the shell now

undergoes an adiabatic expansion the relation between T and R is :

(1) T e–3R (2) T 1

R (3) T 3

1

R (4) T e–R


Sol.[2] 4U KVT

P= 41 U KT

3 V 3

dQ= 0 for adiabatic process dU dW

4d KVT PdV

33 4 KT

4KVT dT KT dv dv3

dT dv dR

T v R

1T

R

66. An inductor (L =0.03H) and a resistor (R=0.15 k) are connected in series to a battery of 15V EMF in a circuitshown below. The key K

1 has been kept closed for a long time. Then at t =0, K

1 is opened and key K

2 is closed

simultaneously. At t=1ms, the current in the circuit will be : (e 150)

(1) 67 mA (2) 6.7 mA (3) 0.67 mA (4) 100 mA

Sol.[3]tR

L0I I e

3 33 115 10 0.15 10

I 10 e0.15 0.03

= 10–5 e–5

= 51

10150

= 0.67mA

67.* A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is

added to its bob, the time period changes to TM. If the Young's modulus of the material of the wire is Y then

1

Yis

equal to : (g =gravitational acceleration)

(1)

2MT Mg

1T A

(2)

2MT A

1T Mg

(3)

2

M

T A1

T Mg

(4)

2MT A

1T Mg

Sol.[1] LT 2

g

mL L

T 2g


2mT L L

T L

2mT L

1T L

2mT L

1T L

2mT Mg

1T AY

2mT1 A

1Y T mg

68. A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of1 m from the diode is :(1) 2.45 V/m (2) 5.48 V/m (3) 7.75 V/m (4) 1.73 V/m

Sol.[1]2

01

I E C2

202

P 1E C

24 R

20

2PE

4 R C

69. Two coaxial solenoids of different radii carry current in the same direction. Let 2F

be the magnetic force on the

inner solenoid due to the outer one and 2F

be the magnetic force on the outer solenoid due to the inner one. Then:

(1) 1F

is radially inwards and 2F

is radially outwards (2) 1F

is radially inwards and 2F

=0

(3) 1F

is radially outwards and 2F

=0 (4) 1F

= 2F

=0

Sol.[4] Due to inner solenoid magnetic field outside is zero.

70.* Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, theaverage time of collision between molecules increases as Vq, where V is the volume of the gas. The value of q is:

p

v

C

C

(1) 3 5

6

(2)

1

2

(3)

1

2

(4)

3 5

6

Sol.[2] 2

1v t

2 N / V

2

3RT V.T

M 2 N

Vt

T

& for adiabatic change


TV constan t

1

CT

V

1/21

Vt

V

1t V

2

qt V

1q

2

71. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q0 and then

connected to the L and R as shown below :

If a student plots graphs of the square of maximum charge (Q 2Max ) on the capacitor with time(t) for two different

values L1 and L

2 (L

1>L

2) of L then which of the following represents this graph correctly ? (plots are schematic and

not drawn to scale)

(1) (2) (3) (4)

Sol. [4]

For damped oscillationQ= Q

0e–Kt

Q2= 2 2kt0Q e

Here constant K is inversly proportional to L


72.* From a solid sphere of mass M and radius R, a spherical portion of radius R

2 is removed, as shown in the figure.

Taking gravitational potential V=0 at r = , the potential at the centre of the cavity thus formed is : (G= gravitationalconstant)

(1) (2) (3) (4)

Sol.[1] Let us consider full sphere of mass m potential at point P isV

p due to full mass

Vp = 3

Gm

2R

22 R

3R2

Potential due to removed portion of sphere is

p 1

3G m / 8V

2 R / 2

= = 3G m

8R

So net potential is

V= p p 1V V

= 11Gm 3Gm

8R 8R

= Gm

R

73.* A train is moving on a straight track with speed 20 ms–1. It is blowing its whistle at the frequency of 1000 Hz. Thepercentage change in the frequency heard by a person standing near the track as the train passes him is (speedof sound=320 ms–1) close to:(1) 12% (2) 18% (3) 24% (4) 6%

Sol. [1] When train is approaching the person frequency heard is 1320 20

f 1000320

117

f 100016

When train passes the person frequency heard is

2320

f 1000320 20

= 1000 ×16

17

% Change = 1 2

1

f f100

f

11.2% 12%


74.*

Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed againsta wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and thewall is 0.15, the frictional force applied by the wall on block B is :

(1) 80 N (2) 120 N (3) 150 N (4) 100 N

Sol.[2]

1f 20N, N=F

1 2100+f f

100+20 = f2

f2 = 120 N

75.* Distance of the centre of mass of a solid uniform cone from its vertex is z0. If the radius of its base is R and its

height is h then z0 is equal to :

(1) 3h

4(2)

5h

8(3)

23h

8R(4)

2h

4R

Sol.[1] Z0 =

3h

4

76. A rectangular loop of sides 10 cm and 5 cm carrying a current of 12 A is placed in different orientations as shownin the figures below :

(a) (b) (c) (d)

If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop would be in (i)stable equilibrium and (ii) unstable equilibrium ?(1) (a) and (c), respectively (2) (b) and (d), respectively (b) and (c), respectively (4) (a) and (b), respectively

Sol.[2] If B & M

are in same direction then the equilibriumis stable i.e. b,d


77.

In the circuit shown, the current in the 1 resistor is :(1) 0 A (2) 0.13 A, from Q to P (3) 0.13 A, from P to Q (4) 1.3 A, from P to Q

Sol.[2]V 6 V 0 V 9

03 1 5

V= 2

23

So current I = 0.13 from Q to P

78. A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ) on its surface. For this

sphere the equipotential surfaces with potentials have radius R1, R

2, R

3 and R

4

respectively. Then(1) R

1 0 and (R

2 – R

1) > (R

4 – R

3) (2) R

1=0 and R

2 < (R

4 – R

3)

(3) 2R < R4

(4) R1=0 and R

2 > (R

4 – R

3)

Sol.[2,3] Potential inside the uniformly charged sphere is

2 23

KQV 3R r

2R

centre3KQ 3V KQ

V v2R 2 R

2R

R2

34

R R3

4R 4R

79. In the given circuit, charge Q2 on the 2µF capacitor changes as C is varied from1µF to 3µF. Q

2 as a function of 'C'

is given properly by : (f igures are drawn schematically and are not to scale)

(1) (2) (3) (4)


Sol.[1]

Charge on 3F

= 3c

E3 c

Potential difference across 3F

33cE 1

V3 E 3

= cE

3 cSo charge on 2F

2 3CE

Q v 2 2 13 C

For c =1F

21

Q F2

& for C= 2F

2Q 1 F and graph is not linear

80.* A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the ydirection with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collisionis close to :(1) 50% (2) 56% (3) 62% (4) 44%

Sol.[2] By momentum

conservation

ˆ ˆm2vi m 2vj 3mv

2v 2vˆ ˆv i j3 3

2 2v V

3

2 2v v

3

2 2i

1 1 K.E m 2v 2m v

2 2


2 21 1m4v 2mv

2 2

21

6 mv2

2

f1 2 2v

K.E 3m2 3

= 28 1

mv3 2

% Loss = i f

i

K.E K.E56%

K.E

81. Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is µ,a ray, incident at an angle , on the face AB would get transmitted through the face AC of the prism provided :

(1) < sin–1 1 1

sin A sin

(2) > cos–1 1 1

sin A sin

(3) < cos–1 1 1

sin A sin

(4) > sin–1 1 1

sin A sin

Sol.[4]

A–R < C

A–R> 1 1

sin

1 1r A sin 1

1 sin sinr 2

From (1) & (2)

1 1Sinr Sin A sin

1sin 1Sin A sin

1 1 1sin sin A sin


82.* From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cubeabout an axis passing through its center and perpendicular to one of its faces is :

(1) 2MR

16 2(2)

24MR

9 3(3)

24MR

3 3(4)

2MR

32 2

Sol.[2]

2R a 3

a= 2R

3

Mass of cube is =

3

3

m 8R4 3 3R3

= 2M

3

moment of inertia of cube about centre and perpendicular to one face is = 2 2m'a 2m 4R

6 3 63

83. Match List – I (Fundamental Experiment) with List – II (its conclusion) and select the correct option from thechoices given below the list :

List – I List – II

(A) Franck–Hertz Experiment (i) Particle nature of light

(B) Photo–electric experiment (ii) Discrete energy levels of atom

(C) Davison – Germer Experiment (iii) Wave nature of electron

(iv) Structure of atom

(1) (A) – (ii) (B) – (iv) (C) – (iii) (2) (A) – (ii) (B) – (i) (C) – (iii)

(3) (A) –(iv) (B) – (iii) (C) – (ii) (4) (A) – (i) (B) – (iv) (C) – (iii)

Sol.[2] (A) Franek-Wertz experiment

Discrete Energy level

(B) Photoelectric experiment

Particle nature of light

(C) Davison-Germer experiment

Wave nature of electron


84. When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is2.5×10–4 ms–1 If the electron density in the wire is 8×1028 m–3, the resistivity of the material is close to :

(1) 1.6×10–7 m (2) 1.6×10–6 m (3) 1.6×10–5 m (4) 1.6×10–8 m

Sol.[3] I = ne AVd

AV

= neAvd

= d

V

neV

= 28 19 4

5

0.1 8 10 1.6 10 2.5 10

= 1.6×10–5 m

85.* For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against itsdisplacement d. Which one of the following represents these correctly ? (graphs are schematic and not drawn toscale)

(1) (2)

(3) (4)

Sol.[1] K.E. = 1

2 mv2

= 1

2m2 (A2–d2)

P.E. = 1

2 m2d2


86.* Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and40 m/s respectively. Which of the following graph best represents the time variation of relative position of thesecond stone with respect to the first ? (Assume stones do not rebound after hitting the ground and neglect airresistance, take g=10 m/s2) (The figures are schematic and not drawn to scale)

(1) (2)

(3) (4)

Sol.[2]

21

1y 240 10t gt

2

22

1y 240 40t gt

2

2 1y y 30t This is valid till the 1st object reach the ground.

After 8 second 1st object reaches the ground and 2nd object is at initial position after that the relative separa-tion is

22

1y 240 40t gt

2

y1 = 0

y2 –y

1= y

2


87.* A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways:

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat.

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat.

In both the cases body is brought from initial temperature 100°C to final temperature 200°C. Entropy change of thebody in the two cases respectively is :

(1) In2, In2 (2) ln2, 2ln2 (3) 2ln2, 8ln2 (4) ln2, 4ln2

Sol.[1] S= ms n T2/T

1

S= entropy

T= temperature of body

S1 =

200n

100

S1 = n2

S2 =

200n

100

= n2

88. Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimumseparation between two objects that human eye can resolve at 500 nm wavelength is :

(1) 30 µm (2) 100 µm (3) 300 µm (4) 1 µm

Sol.[1] 1.22d

= wavelength

d = diameter of eye

& D

= Separation between the object

D= Distance of object from eye

D

= 1.22

d

= 1.22 D

d

= 30 m


Two long current carrying thin wires, both with current , are held by insulating threads of length L and are inequilibrium as shown in the figure, with threads making an angle '' with the vertical. If wires have mass per unitlength then the value of I is : (g =gravitational acceleration)

(1) 2sin 0

gL

cos

(2) 2 0

gL2 tan

(3) 0

gLtan

(4) sin 0

gL

cos

Sol. [1] T cos =mg

T sin =Fm

m

mgCot

F

0

g

I ICot

2 d

= 20

g2 2Lsin

I

d 2 L Sin

I = 2 sin 0

g L

cos

90. On a hot summer night, the refractive index of air is smallest near the ground and increases with height from theground. When a light beam is directed horizontally, the Huygens' principle leads us to conclude that as it travels,the light beam :

(1) goes horizontally without any deflection (2) bends downwards

(3) bends upwards (4) becomes narrower

Sol. [3]

i.e. Light goes upward

JEE (Main) 2015 Question Paper with Solution Chemistry · Which of the following compounds will...

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Transcript of JEE (Main) 2015 Question Paper with Solution Chemistry · Which of the following compounds will...